Question 21.SE.1: What product is formed when radium-226 undergoes alpha emiss...

Analyze We are asked to determine the nucleus that results when radium-226 loses an alpha particle.

Plan We can best do this by writing a balanced nuclear reaction for the process.

Solve The periodic table shows that radium has an atomic number of 88. The complete chemical symbol for radium-226 is therefore {}_{88}^{226}Ra. An alpha particle is a helium-4 nucleus, and so its symbol is {}_2^4 He. The alpha particle is a product of the nuclear reaction, and so the equation is of the form

{}_{88}^{226}Ra \longrightarrow{}_Z^A X +{}_2^4 He

where A is the mass number of the product nucleus and Z is its atomic number. Mass numbers and atomic numbers must balance, so

226 = A + 4

and

88 = Z + 2

Hence,

A = 222      and          Z = 86

Again, from the periodic table, the element with Z = 86 is radon (Rn). The product, therefore, is {}_{86}^{222}Rn, and the nuclear equation is

{}_{88}^{226}Ra \longrightarrow{}_{86}^{222}Rn +{}_2^4 He